Problem: $ F = \left[\begin{array}{rrr}4 & -2 & -1 \\ -2 & 3 & -1\end{array}\right]$ $ C = \left[\begin{array}{rr}-2 & 0 \\ 5 & 5 \\ 4 & 3\end{array}\right]$ What is $ F C$ ?
Answer: Because $ F$ has dimensions $(2\times3)$ and $ C$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ F C = \left[\begin{array}{rrr}{4} & {-2} & {-1} \\ {-2} & {3} & {-1}\end{array}\right] \left[\begin{array}{rr}{-2} & \color{#DF0030}{0} \\ {5} & \color{#DF0030}{5} \\ {4} & \color{#DF0030}{3}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ F$ , with the corresponding elements in column $j$ of the second matrix, $ C$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ F$ with the first element in ${\text{column }1}$ of $ C$ , then multiply the second element in ${\text{row }1}$ of $ F$ with the second element in ${\text{column }1}$ of $ C$ , and so on. Add the products together. $ \left[\begin{array}{rr}{4}\cdot{-2}+{-2}\cdot{5}+{-1}\cdot{4} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ F$ with the corresponding elements in ${\text{column }1}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{-2}+{-2}\cdot{5}+{-1}\cdot{4} & ? \\ {-2}\cdot{-2}+{3}\cdot{5}+{-1}\cdot{4} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ F$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{-2}+{-2}\cdot{5}+{-1}\cdot{4} & {4}\cdot\color{#DF0030}{0}+{-2}\cdot\color{#DF0030}{5}+{-1}\cdot\color{#DF0030}{3} \\ {-2}\cdot{-2}+{3}\cdot{5}+{-1}\cdot{4} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{4}\cdot{-2}+{-2}\cdot{5}+{-1}\cdot{4} & {4}\cdot\color{#DF0030}{0}+{-2}\cdot\color{#DF0030}{5}+{-1}\cdot\color{#DF0030}{3} \\ {-2}\cdot{-2}+{3}\cdot{5}+{-1}\cdot{4} & {-2}\cdot\color{#DF0030}{0}+{3}\cdot\color{#DF0030}{5}+{-1}\cdot\color{#DF0030}{3}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-22 & -13 \\ 15 & 12\end{array}\right] $